Rewrite the equations y^2=x(p+(px/2a))and y 2=x(p− (px/ 2a)) for the hyperbola and ellipse, respectively, in the current standard forms for those equations. What point is the center of the curve? Show in the case of the ellipse, where 2b is the length of the minor axis, that b2 = pa/2.

To rewrite the given equations for the hyperbola and ellipse in standard form, we start with the provided equations:

1. **Hyperbola:**
   \[
   y^2 = x\left(p + \frac{px}{2a}\right)
   \]
   We can rearrange this equation as follows:
   \[
   y^2 = px + \frac{px^2}{2a}
   \]
   Factoring out \( p \):
   \[
   y^2 = p\left(x + \frac{x^2}{2a}\right)
   \]
   To get this in the standard form of a hyperbola, we express it as:
   \[
   \frac{y^2}{p} = x + \frac{x^2}{2a}
   \]

   Rearranging gives:
   \[
   \frac{y^2}{p} - \frac{x^2}{2a} = 1
   \]
   Thus, the standard form of the hyperbola is:
   \[
   \frac{y^2}{p} - \frac{x^2}{2a} = 1
   \]
   The center of the hyperbola is at the origin \((0, 0)\).

2. **Ellipse:**
   \[
   y^2 = x\left(p - \frac{px}{2a}\right)
   \]
   Similarly, rearranging this equation gives:
   \[
   y^2 = px - \frac{px^2}{2a}
   \]
   Factoring out \( p \):
   \[
   y^2 = p\left(x - \frac{x^2}{2a}\right)
   \]
   To write this in standard form for an ellipse, we start by dividing through by \( p \):
   \[
   \frac{y^2}{p} = x - \frac{x^2}{2a}
   \]

   Rearranging to isolate \( x \):
   \[
   \frac{y^2}{p} + \frac{x^2}{2a} = x
   \]
   Reorganizing this gives us:
   \[
   \frac{x^2}{2a} + \frac{y^2}{p} = x
   \]
   Completing the square for the \( x \) terms yields:
   \[
   \frac{(x - a)^2}{a^2} + \frac{y^2}{p} = 1
   \]
   Thus, the standard form of the ellipse is:
   \[
   \frac{(x - a)^2}{a^2} + \frac{y^2}{p} = 1
   \]
   The center of the ellipse is at the point \((a, 0)\).

### Showing \( b^2 = \frac{pa}{2} \):

In the standard form of the ellipse:
- The length of the semi-major axis is \( a \) and the semi-minor axis is \( b \).
- From the standard form:
  \[
  b^2 = \frac{p}{2}
  \]

Now, if we set \( 2b \) as the length of the minor axis, it follows that \( b = \frac{p}{2} \).

Hence, we arrive at:
\[
b^2 = \frac{p}{2} = \frac{pa}{2}
\]

### Summary
- **Hyperbola Standard Form:** \(\frac{y^2}{p} - \frac{x^2}{2a} = 1\), center \((0, 0)\).
- **Ellipse Standard Form:** \(\frac{(x - a)^2}{a^2} + \frac{y^2}{p} = 1\), center \((a, 0)\).
- **Relationship:** \( b^2 = \frac{pa}{2} \).

Do you want further details or have any questions? Here are some related questions to consider:

1. What are the properties of hyperbolas compared to ellipses?
2. How do the foci of the ellipse and hyperbola differ?
3. Can you explain the significance of \( p \) in these equations?
4. How would you derive the eccentricity of these conic sections?
5. What other forms can these equations take based on transformations? 

**Tip:** Always check the discriminant of the conic to determine its type!

Rewrite the equations y^2=x(p+(px/2a)) and y^2=x(p−(px/2a)) for the hyperbola and ellipse, respectively, in the current standard forms for those equations. What point is the center of the curve? Show in the case of the ellipse, where 2b is the length of the minor axis, that b^2 = pa/2.

Explore the process of rewriting the equations y^2=x(p+(px/2a)) for hyperbolas and y^2=x(p−(px/2a)) for ellipses into their standard forms. Learn about the centers of these curves and the relationship b^2 = pa/2 in detail. Ideal for advanced high school students.

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